Chemistry Spm Module Form 4 Chapter 3

8/11/2019 Chemistry Spm Module Form 4 Chapter 3

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CHAPTER 3: CHEMICAL FORMULAE AND

EQUATIONS

Concept of moles

1. One mole is the amount of substance which contains the same number of particles

as there are in 12 grams of carbon-12

2. The number of atoms in 12 grams of carbon-12 is 6.02 X 1023

3. This 6.02 X 1023 is called Avogadro's number or Avogadro's constant (NA)

4. Particles can be

a) atoms,

b) ions

c) molecules

5. Molar mass and relative molecular mass is same meaning

6.

Calculation molar mass (RAM refer to periodic table)

a) What is the molar mass for glucose, C6H12O6? [RAM of C=12, H=1, O= 16]

(12x6) + (1x12) + (16x6)

=180

b) The molecular mass ofM(OH)2is 98. What is the relative atomic mass of atom

M [RAM of H=1, O= 16]

M+ (16+1)2 = 98

M= 9834

M = 64

c) Molar mass of Mn(NO3)2 ? [RAM of Mn=55, N=14, O= 16]

55 + (14x2) + (16x6)

= 179


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7. Calculate the mass (RAM refer to periodic table):

If number of mole is given

and rememberRMM/molar mass always can be found

use above formula to find mass by number mole multiply with RMM

a) 0.05 mol of KMnO4 [answer: 7.9g]

b) 0.1 mol CuSO4 [answer: 16g]

c)

0.08 mol of ascorbic acid, C6H8O6 [answer: 14.08g]

8. Calculate the number of mole for: (RAM refer to periodic table)

If number of mass is given


use above formula to find number mole by mass divide with RMM

a) 2.8g iron [answer: 0.05 mol]

b) 4.05g of nicotine, C10H14N2[answer: 0.025 mol]

c) 1.49g of (NH4)3PO4[answer: 0.01 mol]

d) 2.3g of ethanol, C2H5OH [answer: 0.05 mol]

e) 23.5g of copper(II) nitrate, Cu(NO3)2 [answer: 0.125 mol]


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9. Calculate the number of particles in: (RAM refer to periodic table)

[(Assume NA= 6.02 x 1023mol-1]

Particles can be

i. atoms,

ii. ions

iii. molecules

Based on above formula to find number particles must have number of

mole and Avogadrosconstant, NA


find number of mole first

a) 0.75 mol of Aluminum atoms, Al [answer: 4.52 x 1023Aluminum atoms]

Number of moles x NA

= 0.75 X 6.02 x 1023

= 4.52 x 1023Aluminum atoms (atoms also number of particles)

b) 1.2 mol of chloride ions, Cl- [answer: 7.22 x 1023chloride ions]


= 1.2 X 6.02 x 1023

= 7.22 x 1023chloride ions (ions also number of particles)

c) 0.07 mol of carbon dioxide molecules, CO2[answer: 4.21 x 1022carbon

dioxide molecules


= 0.07 X 6.02 x 1023

= 4.21 x 1022carbon dioxide molecules (molecules also number of particles)


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10.Calculate the number of atoms in: (RAM refer to periodic table)

[Assume NA= 6.02 x 1023mol-1]

Given is molecules but they asked number atoms

Find number of particles first

Find number of atoms in that molecules

Multiply number of particle with the atoms

a) 0.2 mol of sulphur dioxide gas, SO2 [answer: 3.61 x 1023atoms]

Number paticles = 0.2 x 6.02 x 1023

= 1.204 x 1023

Number of atoms in SO2 = 1 Sulphur and 2 Oxygen

= 3 atoms

Number of atoms = 1.204 x 1023 x 3

= 3.61 x 1023atoms

b) 0.125 mol of methane gas CH4 [answer: 3.76 x 1023atoms]

Number paticles = 0.125 x 6.02 x 1023

= 7.525 x 1022

Number of atoms in CH4 = 1 Carbon and 5 hydrogen

= 5 atoms

Number of atoms = 7.525 x 1022 x 5

= 3.76 x 1023atoms


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11.Calculate the number of particles in: (RAM refer to periodic table)

a) 12.8g of copper [answer: 1.2x 1023atoms]

b) 8.5g of ammonia, NH3 [answer: 3.01 x 1023atoms]

c) 2g of hydrogen gas, H2[answer: 6.02 x 1023atoms]

d) 5g of nitrogen dioxide gas NO2 [answer: 6.54 x 1022atoms ]

e) 6.6g of ethane C2H6 [answer: 1.32 x 1023atoms]

(Assume NA= 6 x 1023 mol-1)

Relationship between Number of moles of a gas and its

volume

1. 1 molof any gas at room temperature (250C)and pressure of 1 atmosphereoccupies

24 dm3 (24 000 cm3)

2. At standard temperature and pressure (s.t.p)which at 00C and 1 atmosphere, 1 mol

of gas occupies 22.4 dm3(22 400 cm3)

3.

1 dm3= 1000 cm3


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4. Calculate the volume of: (RAM refer to periodic table, s.t.p= 22.4dm3, r.t.p = 24.0 dm3)

Based on above formula, to find volume, number mole is required

Multiply with stp or rtp based on the question [always given]

a) 0.75 mol nitrogen gas at s.t.p [answer: 16.8 dm3]

b) 0.55 mol of oxygen gas at room temperature and pressure [answer: 13.2 dm3]

5. Calculate the no. of moles of the following gases at room temperature and pressure

(RAM refer to periodic table, s.t.p= 22.4dm3, r.t.p = 24.0 dm3)

Based on above formula, to find number mole, volume is required

Multiply with stp or rtp based on the question

a) 4.8 dm3of chlorine gas [answer: 0.2 mol]

b) 1200 cm3of methane gas [answer: 0.05 mol]

6. Calculate the number of moles of:


a.

672 cm3

carbon dioxide at s.t.p [answer: 0.03 mol] (convert to dm3

first)b. 50 dm3oxygen gas at s.t.p and r.t.p [answer: 2.2 mol, 2.1 mol]

7. Calculate the volume occupied by:


a) Based on above formula, to find volume, MUST GO PASS NUMBER OF

MOLE

b) Find number mole first

c) Then, Multiply with stp or rtp based on the question

a. 1.4g of ethene gas, C2H4at s.t.p [answer: 1.12 dm3]

b. 2.5g of oxygen gas at room temperature and pressure [answer: 1.88 dm3]

c. 4g of carbon dioxide gas at room temperature and pressure [answer: 2.09 dm3]


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8. Calculate the mass of the following


a) Based on above formula, to find mass, MUST GO PASS NUMBER OF

MOLE

b) Find number mole first

c) Then, Multiply with RMM

a. 16.8 dm3of methane gas CH4at s.t.p [answer: 12g]

b. 6720 cm3of carbon monoxide gas, CO at s.t.p [answer: 8.4g]

Empirical and molecular formula

1.Empirical formula is the simplest ratio of the atoms of the

elements that combine to form compound

2.

Molecular formula is the actual numbers of the atoms of the

elements to form the compound

REMEMBER: IF YOU GET SIMPLEST RATIO SUCH AS 0.5, 1.5, 2.5, 3.5,

Please make as whole number by multiplying with 2. DONT EVER LEAVE

IT IN DECIMAL PLACES

IN Calculating empirical formula

1. Write mass or percentage

2. Calculate no of mole

3. Divide with smallest number of mole

4. Get simplest ratio


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3. A hydrocarbon has 85.7% of carbon. The rest are hydrogen. The molar mass is 28

a. (RAM: H=1; C=12)

b. Determine the empirical formula

c. Determine the molecular formula

Empirical formula is CH2

Molecular formula is (CH2)n= molar mass

[12 + (1 x 2)]n= 28

14n = 28

n = 2

Replace n =2 into (CH2)n, the molecular formula is C2H4

4. 2.5g ofXcombined with 4g of Yto form a compound with formula XY2 . If the RAM of

Yis 80, determine the relative atomic mass ofX

5. 1.04g of elementX reacted with 0.48g of oxygen to form an oxide with empirical formula

X2O3.Determine the relative atomic mass of X

Element C HMass (step 1) 85.7% 100- 85.7 = 14.3

Number of

moles

(step 2)

85.7

12

= 7.1 mol

14.3

1

=14.3 mol

Simplest ratio

(step 3)

7.1

7.1= 1

14.3

7.1= 2


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CRUCIBLE LID METHOD IS SUITABLE FOR MAGNESIUM BECAUSE IT IS HIGHER IN

ECS AND VERY REACTIVE COMPARED TO COPPER WHICH IS LESS REACTIVE AND

USES REDUCTION BY HYDROGEN

Crucible lid need to open from time to time to allow oxygen gas to enter and to

react

the hydrogen gas must be passed through first before experiment to remove

all the air in the combustion tube

mixture of air and hydrogen can cause explosion when ignited

weight should be measured from time to time until constant weight is

obtained to assume the reaction completed

hydrogen gas is continuously flow through during cooling process to prevent

any gas to enter again and react with the metal

this causes constant weight could not achieved

Calcium chloride to dry the hydrogen gas

HCl/ H2SO4and Zinc to produce hydrogen gas

CaCl

Reduction by hydrogen method

Crucible lid method


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SOLVING CHEMISTRY QUESTION RULES

1.

Write chemical equation

2.

Balance the equation3.

Find number of mole of any compound that information

given

4.Dont calculate the excess compounds

(Relative Atomic Mass PLEASE REFER TO PERIODIC TABLE)


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QUESTIONS

1. If 0.12g of magnesium reacts with hydrochloric acid, calculate

a. The mass of magnesium chloride formed

b. The volume of hydrogen gas evolved at room temperature and pressure

c.

(RAM; Mg=24, Cl=35.5; 1 mol of gas occupies a volume of 24 dm3at rtp)

Step 1 (Equation):Mg + HClMgCl2 + H2

Step 2 (Balance the equation):Mg + 2HClMgCl2 + H2

Step 3 (Find number of mole of any compoundthat information given):

Information given is for magnesium which has mass and RAM so,

Number of mole = Mass/RAM 0.12g / 24 = 0.005 mole

a) Already found which is 0.005 mole

b) Based on the equation, Mg + 2HClMgCl2 + H2

So number of mole of hydrogen gas is 0.005 mol,

0.005 mol x 24 dm3(given) = 0.12 dm3

Mg2+ Cl-

MgCl2

0.005 0.01 0.005 0.005


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2. 3.2g of copper(II) oxide powder reacted with excess of nitric acid

a. Cu0 + HNO3Cu(NO3)2 + H2O

b. Calculate the mass of copper(II) nitrate formed

c. (RAM; N=14, O=16, Cu=64)

3.

Ethanol burns in air as below

a. C2H5OH +3O22CO2 + 3H2O

b. Calculate the mass of ethanol burnt if 2.4dm3of carbon dioxide is produced at

room temperature

c. (RAM; H=1, C=12, O=16; 1 mol of gas occupies a volume of 24 dm3at rtp)

4. A hydrocarbonXcontains 82.76% carbon by mass. 2.9g of hydrocarbonX occupies a

volume of 1.2 dm3at room temperature and pressure. (RAM: H=1, C=12, O=16; 1 mol of

gas occupies a 24dm3at rtp;NA=6 X 1023mol-1)

a.

Determine the empirical formula of hydrocarbonXb. Determine the molecular formula of hydrocarbonX

c. Combustion of X in air produces carbon dioxide and water. Write chemical

equation

d. If 11.6g of X burnt

e. The mass of water formed

f. The number of carbon dioxide molecules produced at room temperature

5. K + O2K2O

a. Calculate the mass number of mole for potassium oxide if 23.5g of potassium

oxide is usedb. Calculate the number of mole of potassium used

c. Calculate the mass of potassium used

d. Calculate the volume of oxygen used

e. (STP: 22.4dm3)

6. H2O2H2O + O2

a. (rtp: 24dm3)

b. Calculate the number mole of oxygen produced if 11.2dm3oxygen produced

c. Calculate the number of mole of H2O2 used

d.

Calculate the mass of H2O2 used

7. 8.0g of copper (II) oxide powder is added to excess dilute nitric acid and heated. Find the

mass of copper (II) nitrate formed

8. 1.3g of zinc reacts with excess dilute sulphuric acid. The product formed are zinc

sulphate and hydrogen gas. Find the volume of hydrogen released at STP


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9. C3H8 + O2CO2+ H2O

a. 720 cm3of propane gas at rtp burns in excess oxygen. Calculate the mass of

carbon dioxide formed

10.Hydrogen peroxide decomposes according to the following equation:

222

OH (l)2 OH2

(l) +2O (g)

a. Calculate the volume of oxygen gas,2O measured at STP that can be obtained

from the decomposition of 34 g of hydrogen peroxide,22

OH . [Relative atomic

mass : H, 1 ; O, 16. Molar volume : 22.4 3dm 1mol at STP] (Ans:

11.2 dm3)

11.Silver carbonate Ag2CO3breaks down easily when heated to produce silver metal

2 Ag2CO3(l) 4Ag(s) + 2 2CO (g) + 2O

a. Find the mass of silver carbonatethat is required to produce 10 g of silver

[Relative atomic mass: C, 12 ; O, 16 ; Ag, 108]

(Ans : 12.77g)

12.16 g of copper (II) oxide, CuO is reacted with excess methane,4

CH . Using the equation

below, find the mass of copper that is produced.

[Relative atomic mass: Cu, 64; O, 16]

4 CuO(s) +4

CH (g)4 Cu (s) +2

CO (g) + 2 OH2

(l)

(Ans : 12.8 g)

13. A student heats 20 g of calcium carbonate3

CaCO strongly. It decomposes according to

the equation below:

3CaCO (s) CaO (s) +

2CO (g).

(a). If the carbon dioxide produced is collected at room conditions, what is its volume?

(b). Calculate the mass of calcium oxide, CaO produced.

[Relative atomic mass: C, 12 ; O, 16; Ca, 40. Molar volume : 24 dm3 1mol at

room conditions]

(Ans : (a). 4.8 dm 3(b) 11.2 g)

Chemistry Spm Module Form 4 Chapter 3

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